   Chapter 15, Problem 9P

Chapter
Section
Textbook Problem

(a) Show that when Laplace’s equation ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 + ∂ 2 u ∂ z 2 = 0 is written in cylindrical coordinates, it becomes ∂ 2 u ∂ r 2 + 1 r ∂ u ∂ r + 1 r 2 ∂ 2 u ∂ θ 2 + ∂ 2 u ∂ z 2 = 0 (b) Show that when Laplace’s equation is written in spherical coordinates, it becomes ∂ 2 u ∂ ρ 2 + 2 ρ ∂ u ∂ ρ + cot ϕ ρ 2 ∂ u ∂ ϕ + 1 ρ 2 sin 2 ϕ ∂ 2 u ∂ θ 2 = 0

(a)

To determine

To show: The Laplace equation in cylindrical coordinates is 2ur2+(1r)ur+(1r2)2uθ2+2uz2=0.

Explanation

Given:

The Laplace equation is 2ux2+2uy2+2uz2=0.

Formula used:

The cylindrical coordinates (r,θ,z) corresponding to the rectangular coordinates (x,y,z) is,

r=x2+y2θ=tan1(yx)z=z (1)

The spherical coordinates (ρ,θ,ϕ) corresponding to the rectangular coordinates (x,y,z) is,

ρ=x2+y2+z2ϕ=cos1(zρ)θ=cos1(xρsinϕ) (2)

Chain Rule:

“Suppose that z=f(x,y,z) is a differentiable function of x, y and z where x=g(t),y=h(t),z=k(t) are both differentiable functions of t then, u is differentiable function of t and ut=uxxt+uyyt+uzzt”.

Calculation:

From the chain rule mentioned above, it is observed that ur=uxxr+uyyr+uzzr. From the equation (1), it is observed that x=rcosθ,y=rsinθ,z=z. So,

xr=cosθ,yr=sinθ,zr=0

Substitute this in the previous equation.

ur=uxxr+uyyr+uzzr=uxcosθ+uysinθ+uz(0)=uxcosθ+uysinθ

Differentiate this again.

r(ur)=r(uxcosθ+uysinθ)={cosθ[2ux2(xr)+2uyx(yr)+2uzx(zr)]+sinθ[2uy2(yr)+2uxy(xr)+2uzy(zr)]}={cosθ[2ux2(cosθ)+2uyx(sinθ)+2uzx(0)]+sinθ[2uy2(sinθ)+2uxy(cosθ)+2uzy(0)]}=2ux2cos2θ+2uy2sin2θ+22uyxsinθcosθ

Thus, 2ur2=2ux2cos2θ+2uy2sin2θ+22uyxsinθcosθ.

Again from the chain rule mentioned above, it is observed that uθ=uxxθ+uyyθ+uzzθ. From the equation (1), it is observed that x=rcosθ,y=rsinθ,z=z. So,

xθ=rsinθ,yθ=rcosθ,zθ=0

Substitute this in the previous equation.

uθ=uxxθ+uyyθ+uzzθ=ux(rsinθ)+uy(rsinθ)+uz(0)=uxrsinθ+uyrsinθ

Differentiate this again

(b)

To determine

To show: The Laplace equation in spherical coordinates is 2uρ2+(2ρ)uρ+(cotϕρ2)uϕ+(1ρ2)2uϕ2+(1ρ2sin2ϕ)2uθ2=0.

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